Galois Group of Reducible Polynomial is Direct Product of Galois Groups of Factors?

I have a question about Galois groups of reducible separable polynomials. For example x^5+x+1 = (x^2+x+1)(x^3-x^2+1). We can see by direct computation that it's solvable and not contained in A5. The first factor has Gal = Z/2, and the other one, according to the program GAP, has Gal of order 2, so S2. My question then, what is Gal for x^5+x+1, is it S2 x S2? If so, is it the case for f with factorization p1...p_n and G:=Gal(p_i) that Gal(f) = G1 x ... x G_n **(nope)**? Thank you.

edit: I might just throw in this stupid question as well: If all the factors of f=p1...p_n has solvable Galois group, is f also solvable?

Re: Galois Group of Reducible Polynomial is Direct Product of Galois Groups of Factor

last question first:

if all the factors of f(x) have solvable galois groups, then those factors themselves are solvable (by radicals), so f is solvable by radicals.

now let's look at the more general situation:

you are asking: if f(x) = p(x)q(x) in F[x], and if Gal(p) = H, Gal(q) = K, is Gal(f) ≅ HxK?

the thing is, Gal(f) may have a subgroup isomorphic to H, and a subgroup isomorphic to K, but these may "overlap" (even if f is separable).

consider f(x) = (x^{2} - 2)(x^{2}- 2x - 1) the splitting field for BOTH factors is Q(√2) so the galois group is only of order 2, not of order 4.

Re: Galois Group of Reducible Polynomial is Direct Product of Galois Groups of Factor

Thank you. So how does one work with possible overlaps, taking intersection? I'm not sure how I would extract any information from that.

Also, I don't know what GAP (or I) was thinking with regards to x^3-x^2+1 above; it's irreducible and thus has Galois group S3 (its discriminant is not a square). According to a work by Valibouze, if we set f=x^5+x+1, p=x^2+x+1, q=x^3-x^2+1, Gal(f) is contained in Gal(p) x Gal(q). More precisely, somehow, Gal(f) = <(345),(12)(45)>. My question then (which hardly qualifies as advanced algebra, but it's unnecessary to start a new thread), how do I find the elements of this group and hence its order? If a=(345), b=(12)(45), I've found the following elements:

a^2 = (354) = bab

ab = (12)(34) = ba^2

ba = (12)(35) = a^2 b

so six elements including the identity. I have no idea how many "combinations" of these I need to multiply in order to be "done"?

One more question: Consider the polynomial x^3-2. In every textbook I've read that brings up this example (they all do), the splitting field is obtained by adjoining (-1+sqrt(-3))/2 (and 2^(1/3) of course). My question then: Why not just adjoin sqrt(-3)? All the roots can be obtained by doing field arithmetic on sqrt(-3) and 2^(1/3).

Edit: Groups of order 6 - Groupprops Apparently the above doubly generated group, if indeed it has six elements, is isomorphic to S6. That doesn't seem right?

Re: Galois Group of Reducible Polynomial is Direct Product of Galois Groups of Factor

it's hard to know what group you have just from knowing its subgroups. part of this is because groups can be built in different ways even given the same "building blocks".

with regard to the group <(3 4 5),(1 2)(4 5)> of S_{5}, the most important relation you have is:

ba = a^{2}b.

this tells us any element in <a,b> (which is of the form a^{k1}b^{k2}......a^{kn-1}b^{kn} for

integers k_{1},...,k_{n}) can be put in the form a^{j}b^{k} (we use ba = a^{2}b to "put the a's in front").

since a is of order 3, and b is of order 2, that gives at most 6 elements of <a,b>. since {e,a,a^{2},b,ab,a^{2}b} are 6 distinct elements:

{e, (3 4 5), (3 5 4), (1 2)(4 5), (1 2)(3 4), (1 2)(3 5)}, this is all of them.

that this is isomorphic to S_{3} can be seen in 3 ways:

1) using the isomorphism:

e ---> e

(3 4 5) ---> (1 2 3)

(3 5 4) ---> (1 3 2)

(1 2)(4 5) ---> (1 2)

(1 2)(3 4) ---> (1 3)

(1 2)(3 5) ---> (2 3)

2) noting that if we call (1 2 3) "A", and (1 2) "B" that S_{3} = <A,B> where A^{3} = B^{2} = e, and BA = A^{2}B

thus a--->A, b--->B yields an isomorphism.

3) there are just two groups (up to isomorphism) of order 6, Z_{6} and S_{3}. your group contains no element of order 6, and is thus not cyclic.

i don't know which work (by Valibouze) you are referring to, but the above group is NOT Gal(p) x Gal(q). perhaps you are referring to the SEMI-direct product, and not the direct product

(Gal(q) x| Gal(p) , not Gal(p) x Gal(q)).

********

the elements we adjoin to get an extension field are not "unique". for example Q(1+√2) is the same field as Q(√2).

in general, to get all n-th roots of a rational number r > 0, we need not only the real n-th root, r^{1/n}, but also all n-th roots of unity in C

(square roots are special, because the rationals already contain all square roots of 1). it really doesn't matter if we do this by adjoining:

a)a primitive n-th root ρ of unity to Q(r^{1/n})

b)the n-th root r^{1/n}ρ of r to Q(r^{1/n}).

when n = 3, the primitive cube roots of unity are:

{-1/2 + (√3/2)i, -1/2 - (√3/2)i}. these are often denoted as ω and ω*.

it should be clear that Q(2^{1/3},ω) = Q(2^{1/3},(2ω+1))

clearly 2ω+1 is in Q(2^{1/3},ω), and ω = (1/2)(2ω+1) - 1/2, so ω is in Q(2^{1/3},(2ω+1)).

it may not be obvious that ω^{3} = 1. this is usually done by showing ω^{2} + ω + 1 = 0

(since x^{2} + x + 1 = (x^{3} - 1)/(x - 1)).

note that ω^{2} = (-1/2 + (√3/2)i)^{2} = (1/4 - 3/4) - 2(√3/4)i = -1/2 - (√3/2)i = ω*

so ω^{2} + ω + 1 = ω + ω* + 1 = 2Re(ω) + 1 = 2(-1/2) + 1 = -1 + 1 = 0.

this also makes it clear that ω* = ω^{2} is the other root of x^{2} + x + 1:

x^{3} - 1 = (x - 1)(x - ω)(x - ω*)

(roots occur in "conjugate pairs").

it is easy to verify that:

x^{3} - 2 = (x - 2^{1/3})(x - 2^{1/3}ω)(x - 2^{1/3}ω*)

note that the galois group of x^{3} - 2 has an automorphism which sends ω--->ω* of order 2, and an automorphism which sends 2^{1/3}-->2^{1/3}ω, and sends 2^{1/3}ω ---> 2^{1/3}ω^{2}, which is of order 3, and since:

[Q(2^{1/3},ω):Q] = [Q(2^{1/3},ω):Q(2^{1/3})][Q(2^{1/3}):Q] = 2*3 = 6

we have a galois group isomorphic to S_{3} again, since these automorphisms do not commute. we can write these six automorphisms as:

ω-->ω

2^{1/3}--->2^{1/3} (the identity)

ω-->ω*

2^{1/3}--->2^{1/3} (conjugation)

ω-->ω

2^{1/3}--->2^{1/3}ω (cyclic permutation of roots)

ω-->ω

2^{1/3}--->2^{1/3}ω* (the other cyclic permutation of roots)

ω-->ω*

2^{1/3}--->2^{1/3}ω (fixes 2^{1/3}ω*)

ω-->ω*

2^{1/3}--->2^{1/3}ω* (fixes 2^{1/3}ω).

Re: Galois Group of Reducible Polynomial is Direct Product of Galois Groups of Factor

1-2) Thank you.

3) No no, I said Gal(pq) is contained in Gal(p) x Gal(q), which in this case is S2 x S3 ~= D6 of order 12. Just as e.g Gal(x^5+1) is contained in but not equal to S5, I'm not saying Gal(pq) is equal to Gal(p) x Gal(q), only contained in it, namely the group which you showed isomorphic to S3. The article by Valibouze is "Galois Theory and Reducible Polynomials", if you're interested.

4) I may just be completely misunderstanding you here, but it looks like you're constructing spl(x^3-2)? Let me restate and be more specific: Why adjoin (-1+sqrt(-3))/2 and not just sqrt(-3)? Let u = 2^(1/3), v = (-3)^(1/2) and consider span{1,u,u^2,v,uv,u^2 v}. Then all the roots of x^3-2 can be obtained by choosing appropriate "coordinates", namely 2^(1/3) = (0,1,0,0,0,0), 2^(1/3) (-1+v)/2 = (0,1,0,0,0,0)*(-1/2,0,0,1/2), etc. (with appropriate multiplication)

Re: Galois Group of Reducible Polynomial is Direct Product of Galois Groups of Factor

3) my bad, for some reason i was thinking Gal(q) was C_{3}, not S_{3}, and S_{2} x C_{3} only has order 6.

it turns out that the group you exhibited is indeed in S_{2} x S_{3}, note that Gal(pq) acts independently on the orbits {1,2} and {3,4,5}.

this will work if the roots of p and the roots of q lie in extensions whose intersection is just Q (this is the "overlap" problem i referred to earlier).

mere separability alone is not enough to guarantee this, however.

4) galois groups (of polynomials f(x) in F[x]) ARE (field) automorphisms of the splitting field of f that fix F. that is how we get the splitting fields in the first place, adjoining roots.

why adjoin [-1+√(-3)]/2? because that is "natural"...it is a root of x^{3} - 1 not contained in Q.

yes, any two vector spaces of dimension 6 over Q are isomorphic, but we also want to preserves "the subspace" (intermediate field) structure, as well.

it is easier to describe a galois field by its actions on the generators of the splitting field directly (which in turn then becomes a permutation groups on the roots).

but yes, we CAN adjoin i√3 instead, we get the same field. it's just that i√3 is not a root of x^{3}-2, or "directly" related to this polynomial (the only way you would KNOW to adjoin i√3 is if you already knew "which complex number" was a cube root of unity. and to describe the galois group we don't really need the real and imaginary coefficients of this complex number, it's enough just to call it ω).

a similar situation exists with the fourth roots of 2 (x^{4} - 2). we can adjoin a real (positive or negative) fourth root of 2 and then i (a primitive fourth root of 1) OR:

we can adjoin a real fourth root of 2, and (2^{1/4})i. it really doesn't matter which one we do.

Re: Galois Group of Reducible Polynomial is Direct Product of Galois Groups of Factor

3) Wait, now I'm confused. Are you saying that Gal(x^5+x+1) = S2 x S3? (~= D6). I have Gal(x^5+x+1) = <(345),(12)(45)>, which you showed to be isomorphic to S3.

Edit: Here's a new question: Gal(x^4-4x^2+2) = V = Z2 x Z2. I have a number of methods at my disposal to show that, but how does one show this by looking at permutations? Say I want to have a permutation that takes sqrt(2+sqrt(2)) to sqrt(2-sqrt(2)) and leaves the other roots invariant, what kind of rule is that? "change the 'inner' sign", that only sounds arbitrary. (ok, that particular permutation doesn't accomplish what I want, but I think you get my point)